ANSWER on Question #78511 – Math – Differential Equations
QUESTION
Solve
∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ x 2 \frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \cdot \frac {\partial^ {2} U}{\partial x ^ {2}} ∂ x 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U
1. U ( x , y ) = cos ( 3 x − y ) U(x,y) = \cos (3x - y) U ( x , y ) = cos ( 3 x − y )
2. U ( x , y ) = x 2 + y 2 U(x,y) = x^{2} + y^{2} U ( x , y ) = x 2 + y 2
3. U ( x , y ) = sin ( 3 x − y ) U(x,y) = \sin (3x - y) U ( x , y ) = sin ( 3 x − y )
4. U ( x , y ) = e − 3 / p i x ⋅ sin ( π y ) U(x,y) = e^{-3 / pix}\cdot \sin (\pi y) U ( x , y ) = e − 3/ p i x ⋅ sin ( π y )
SOLUTION
Remarks related to the condition of the problem:
1. We see that the answers are given as functions of two variables - U ( x , y ) U(x, y) U ( x , y )
1. ∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ y 2 \frac{\partial^2 U}{\partial x^2} = 9 \cdot \frac{\partial^2 U}{\partial y^2} ∂ x 2 ∂ 2 U = 9 ⋅ ∂ y 2 ∂ 2 U ∂ 2 U ∂ y 2 = 9 ⋅ ∂ 2 U ∂ x 2 \frac{\partial^2 U}{\partial y^2} = 9 \cdot \frac{\partial^2 U}{\partial x^2} ∂ y 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U
For each of these cases, we check the solutions provided.
2. Since there are no boundary conditions, we can not apply the method of separation of variables and reduce this problem to the Sturm-Liouville problem.
3. Variant # 4 for the formula editor looks like
U ( x , y ) = e − 3 / p i x ⋅ sin ( π y ) → U ( x , y ) = e − 3 π x ⋅ sin ( π y ) U (x, y) = e ^ {- 3 / p i x} \cdot \sin (\pi y) \rightarrow U (x, y) = e ^ {- \frac {3}{\pi x}} \cdot \sin (\pi y) U ( x , y ) = e − 3/ p i x ⋅ sin ( π y ) → U ( x , y ) = e − π x 3 ⋅ sin ( π y )
But I think that should be so
U ( x , y ) = e − 3 / p i x ⋅ sin ( π y ) → U ( x , y ) = e − 3 π x ⋅ sin ( π y ) U (x, y) = e ^ {- 3 / p i x} \cdot \sin (\pi y) \rightarrow U (x, y) = e ^ {- 3 \pi x} \cdot \sin (\pi y) U ( x , y ) = e − 3/ p i x ⋅ sin ( π y ) → U ( x , y ) = e − 3 π x ⋅ sin ( π y )
Therefore, two cases will also be considered.
We will use this method of solution: we calculate all the partial derivatives and substitute them in the original equation. If equality holds, then it is a solution for us.
1 CASE:
∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ y 2 \frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \cdot \frac {\partial^ {2} U}{\partial y ^ {2}} ∂ x 2 ∂ 2 U = 9 ⋅ ∂ y 2 ∂ 2 U
1. U ( x , y ) = cos ( 3 x − y ) U(x,y) = \cos (3x - y) U ( x , y ) = cos ( 3 x − y )
U ( x , y ) = cos ( 3 x − y ) → { ∂ U ∂ x = − 3 sin ( 3 x − y ) ∂ 2 U ∂ x 2 = − 9 cos ( 3 x − y ) ∂ U ∂ y = sin ( 3 x − y ) ∂ 2 U ∂ y 2 = − cos ( 3 x − y ) U (x, y) = \cos (3 x - y) \rightarrow \left\{\begin{array}{l}\frac {\partial U}{\partial x} = - 3 \sin (3 x - y)\\\frac {\partial^ {2} U}{\partial x ^ {2}} = - 9 \cos (3 x - y)\\\frac {\partial U}{\partial y} = \sin (3 x - y)\\\frac {\partial^ {2} U}{\partial y ^ {2}} = - \cos (3 x - y)\end{array}\right. U ( x , y ) = cos ( 3 x − y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = − 3 sin ( 3 x − y ) ∂ x 2 ∂ 2 U = − 9 cos ( 3 x − y ) ∂ y ∂ U = sin ( 3 x − y ) ∂ y 2 ∂ 2 U = − cos ( 3 x − y )
Then,
{ ∂ 2 U ∂ x 2 = − 9 cos ( 3 x − y ) 9 ⋅ ∂ 2 U ∂ y 2 = 9 ⋅ [ − cos ( 3 x − y ) ] = − 9 cos ( 3 x − y ) → ∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ y 2 \left\{ \begin{array}{c} \frac {\partial^ {2} U}{\partial x ^ {2}} = - 9 \cos (3 x - y) \\ 9 \cdot \frac {\partial^ {2} U}{\partial y ^ {2}} = 9 \cdot [ - \cos (3 x - y) ] = - 9 \cos (3 x - y) \end{array} \right. \to \frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \cdot \frac {\partial^ {2} U}{\partial y ^ {2}} { ∂ x 2 ∂ 2 U = − 9 cos ( 3 x − y ) 9 ⋅ ∂ y 2 ∂ 2 U = 9 ⋅ [ − cos ( 3 x − y )] = − 9 cos ( 3 x − y ) → ∂ x 2 ∂ 2 U = 9 ⋅ ∂ y 2 ∂ 2 U
Conclusion,
U ( x , y ) = cos ( 3 x − y ) − is solution of the given equation U (x, y) = \cos (3 x - y) - \text{is solution of the given equation} U ( x , y ) = cos ( 3 x − y ) − is solution of the given equation
2. U ( x , y ) = x 2 + y 2 U(x,y) = x^{2} + y^{2} U ( x , y ) = x 2 + y 2
U ( x , y ) = x 2 + y 2 → { ∂ U ∂ x = 2 x ∂ 2 U ∂ x 2 = 2 ∂ U ∂ y = 2 y ∂ 2 U ∂ y 2 = 2 U(x, y) = x^{2} + y^{2} \rightarrow \left\{ \begin{array}{l} \frac{\partial U}{\partial x} = 2x \\ \frac{\partial^{2} U}{\partial x^{2}} = 2 \\ \frac{\partial U}{\partial y} = 2y \\ \frac{\partial^{2} U}{\partial y^{2}} = 2 \end{array} \right. U ( x , y ) = x 2 + y 2 → ⎩ ⎨ ⎧ ∂ x ∂ U = 2 x ∂ x 2 ∂ 2 U = 2 ∂ y ∂ U = 2 y ∂ y 2 ∂ 2 U = 2
Then,
{ ∂ 2 U ∂ x 2 = 2 9 ⋅ ∂ 2 U ∂ y 2 = 9 ⋅ 2 = 18 → ∂ 2 U ∂ x 2 = 2 ≠ 18 = 9 ⋅ ∂ 2 U ∂ y 2 \left\{ \begin{array}{c} \frac{\partial^{2} U}{\partial x^{2}} = 2 \\ 9 \cdot \frac{\partial^{2} U}{\partial y^{2}} = 9 \cdot 2 = 18 \end{array} \right. \rightarrow \frac{\partial^{2} U}{\partial x^{2}} = 2 \neq 18 = 9 \cdot \frac{\partial^{2} U}{\partial y^{2}} { ∂ x 2 ∂ 2 U = 2 9 ⋅ ∂ y 2 ∂ 2 U = 9 ⋅ 2 = 18 → ∂ x 2 ∂ 2 U = 2 = 18 = 9 ⋅ ∂ y 2 ∂ 2 U
Conclusion,
U ( x , y ) = x 2 + y 2 − is not solution of the given equation U(x,y) = x^{2} + y^{2} - \text{is not solution of the given equation} U ( x , y ) = x 2 + y 2 − is not solution of the given equation
3. U ( x , y ) = sin ( 3 x − y ) U(x,y) = \sin(3x - y) U ( x , y ) = sin ( 3 x − y )
U ( x , y ) = sin ( 3 x − y ) → { ∂ U ∂ x = 3 cos ( 3 x − y ) ∂ 2 U ∂ x 2 = − 9 sin ( 3 x − y ) ∂ U ∂ y = − cos ( 3 x − y ) ∂ 2 U ∂ y 2 = − sin ( 3 x − y ) U(x, y) = \sin(3x - y) \rightarrow \left\{ \begin{array}{l} \frac{\partial U}{\partial x} = 3 \cos(3x - y) \\ \frac{\partial^{2} U}{\partial x^{2}} = -9 \sin(3x - y) \\ \frac{\partial U}{\partial y} = -\cos(3x - y) \\ \frac{\partial^{2} U}{\partial y^{2}} = -\sin(3x - y) \end{array} \right. U ( x , y ) = sin ( 3 x − y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = 3 cos ( 3 x − y ) ∂ x 2 ∂ 2 U = − 9 sin ( 3 x − y ) ∂ y ∂ U = − cos ( 3 x − y ) ∂ y 2 ∂ 2 U = − sin ( 3 x − y )
Then,
{ ∂ 2 U ∂ x 2 = − 9 sin ( 3 x − y ) 9 ⋅ ∂ 2 U ∂ y 2 = 9 ⋅ [ − sin ( 3 x − y ) ] = − 9 sin ( 3 x − y ) → ∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ y 2 \left\{ \begin{array}{c} \frac{\partial^{2} U}{\partial x^{2}} = -9 \sin(3x - y) \\ 9 \cdot \frac{\partial^{2} U}{\partial y^{2}} = 9 \cdot \left[ -\sin(3x - y) \right] = -9 \sin(3x - y) \end{array} \right. \rightarrow \frac{\partial^{2} U}{\partial x^{2}} = 9 \cdot \frac{\partial^{2} U}{\partial y^{2}} { ∂ x 2 ∂ 2 U = − 9 sin ( 3 x − y ) 9 ⋅ ∂ y 2 ∂ 2 U = 9 ⋅ [ − sin ( 3 x − y ) ] = − 9 sin ( 3 x − y ) → ∂ x 2 ∂ 2 U = 9 ⋅ ∂ y 2 ∂ 2 U
Conclusion,
U(x,y) = \sin (3x - y) - is solution of the given equation
4.A. U ( x , y ) = e − 3 π x ⋅ sin ( π y ) U(x,y) = e^{-3\pi x}\cdot \sin (\pi y) U ( x , y ) = e − 3 π x ⋅ sin ( π y )
U ( x , y ) = e − 3 π x ⋅ sin ( π y ) → { ∂ U ∂ x = − 3 π ⋅ e − 3 π x ⋅ sin ( π y ) ∂ 2 U ∂ x 2 = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ U ∂ y = π ⋅ e − 3 π x ⋅ cos ( π y ) ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) U (x, y) = e ^ {- 3 \pi x} \cdot \sin (\pi y) \rightarrow \left\{\begin{array}{l}\frac {\partial U}{\partial x} = - 3 \pi \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y)\\\frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y)\\\frac {\partial U}{\partial y} = \pi \cdot e ^ {- 3 \pi x} \cdot \cos (\pi y)\\\frac {\partial^ {2} U}{\partial y ^ {2}} = - \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y)\end{array}\right. U ( x , y ) = e − 3 π x ⋅ sin ( π y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = − 3 π ⋅ e − 3 π x ⋅ sin ( π y ) ∂ x 2 ∂ 2 U = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ y ∂ U = π ⋅ e − 3 π x ⋅ cos ( π y ) ∂ y 2 ∂ 2 U = − π 2 ⋅ e − 3 π x ⋅ sin ( π y )
Then,
{ ∂ 2 U ∂ x 2 = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) 9 ⋅ ∂ 2 U ∂ y 2 = 9 ⋅ [ − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ] = − 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) → \left\{ \begin{array}{c} \frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) \\ 9 \cdot \frac {\partial^ {2} U}{\partial y ^ {2}} = 9 \cdot [ - \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) ] = - 9 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) \end{array} \right. \to { ∂ x 2 ∂ 2 U = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) 9 ⋅ ∂ y 2 ∂ 2 U = 9 ⋅ [ − π 2 ⋅ e − 3 π x ⋅ sin ( π y )] = − 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) → ∂ 2 U ∂ x 2 = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ≠ − 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) = 9 ⋅ ∂ 2 U ∂ y 2 \frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) \neq - 9 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) = 9 \cdot \frac {\partial^ {2} U}{\partial y ^ {2}} ∂ x 2 ∂ 2 U = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) = − 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) = 9 ⋅ ∂ y 2 ∂ 2 U
Conclusion,
U(x,y) = e^{-3\pi x} \cdot \sin(\pi y) - is not solution of the given equation
4.B. U ( x , y ) = e − 3 π x ⋅ sin ( π y ) U(x,y) = e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) U ( x , y ) = e − π x 3 ⋅ sin ( π y )
U ( x , y ) = e − 3 π x ⋅ sin ( π y ) → { ∂ U ∂ x = 3 π x 2 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ 2 U ∂ x 2 = − 6 π x 3 ⋅ e − 3 π x ⋅ sin ( π y ) + 9 π 2 ⋅ x 4 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ U ∂ y = π ⋅ e − 3 π x ⋅ cos ( π y ) ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) U(x,y) = e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \rightarrow \left\{ \begin{array}{c}
\frac{\partial U}{\partial x} = \frac{3}{\pi x^2} \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \\
\frac{\partial^2 U}{\partial x^2} = -\frac{6}{\pi x^3} \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) + \frac{9}{\pi^2 \cdot x^4} \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \\
\frac{\partial U}{\partial y} = \pi \cdot e^{-\frac{3}{\pi x}} \cdot \cos(\pi y) \\
\frac{\partial^2 U}{\partial y^2} = -\pi^2 \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y)
\end{array} \right. U ( x , y ) = e − π x 3 ⋅ sin ( π y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = π x 2 3 ⋅ e − π x 3 ⋅ sin ( π y ) ∂ x 2 ∂ 2 U = − π x 3 6 ⋅ e − π x 3 ⋅ sin ( π y ) + π 2 ⋅ x 4 9 ⋅ e − π x 3 ⋅ sin ( π y ) ∂ y ∂ U = π ⋅ e − π x 3 ⋅ cos ( π y ) ∂ y 2 ∂ 2 U = − π 2 ⋅ e − π x 3 ⋅ sin ( π y )
Then,
{ ∂ 2 U ∂ x 2 = ( 9 π 2 x 4 − 6 π x 3 ) ⋅ e − 3 π x ⋅ sin ( π y ) 9 ⋅ ∂ 2 U ∂ y 2 = 9 ⋅ [ − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ] = − 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) → \left\{ \begin{array}{c}
\frac{\partial^2 U}{\partial x^2} = \left(\frac{9}{\pi^2 x^4} - \frac{6}{\pi x^3}\right) \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \\
9 \cdot \frac{\partial^2 U}{\partial y^2} = 9 \cdot \left[ -\pi^2 \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \right] = -9\pi^2 \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y)
\end{array} \right. \rightarrow { ∂ x 2 ∂ 2 U = ( π 2 x 4 9 − π x 3 6 ) ⋅ e − π x 3 ⋅ sin ( π y ) 9 ⋅ ∂ y 2 ∂ 2 U = 9 ⋅ [ − π 2 ⋅ e − π x 3 ⋅ sin ( π y ) ] = − 9 π 2 ⋅ e − π x 3 ⋅ sin ( π y ) → ∂ 2 U ∂ x 2 = ( 9 π 2 x 4 − 6 π x 3 ) ⋅ e − 3 π x ⋅ sin ( π y ) ≠ − 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) = 9 ⋅ ∂ 2 U ∂ y 2 \frac{\partial^2 U}{\partial x^2} = \left(\frac{9}{\pi^2 x^4} - \frac{6}{\pi x^3}\right) \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \neq -9\pi^2 \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) = 9 \cdot \frac{\partial^2 U}{\partial y^2} ∂ x 2 ∂ 2 U = ( π 2 x 4 9 − π x 3 6 ) ⋅ e − π x 3 ⋅ sin ( π y ) = − 9 π 2 ⋅ e − π x 3 ⋅ sin ( π y ) = 9 ⋅ ∂ y 2 ∂ 2 U
Conclusion,
U ( x , y ) = e − 3 π x ⋅ sin ( π y ) − is not solution of the given equation \boxed{U(x,y) = e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) - \text{is not solution of the given equation}} U ( x , y ) = e − π x 3 ⋅ sin ( π y ) − is not solution of the given equation
2 CASE:
∂ 2 U ∂ y 2 = 9 ⋅ ∂ 2 U ∂ x 2 \frac {\partial^ {2} U}{\partial y ^ {2}} = 9 \cdot \frac {\partial^ {2} U}{\partial x ^ {2}} ∂ y 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U
1. U ( x , y ) = cos ( 3 x − y ) U(x,y) = \cos (3x - y) U ( x , y ) = cos ( 3 x − y )
U ( x , y ) = cos ( 3 x − y ) → { ∂ U ∂ x = − 3 sin ( 3 x − y ) ∂ 2 U ∂ x 2 = − 9 cos ( 3 x − y ) ∂ U ∂ y = sin ( 3 x − y ) ∂ 2 U ∂ y 2 = − cos ( 3 x − y ) U (x, y) = \cos (3 x - y) \rightarrow \left\{\begin{array}{l}\frac {\partial U}{\partial x} = - 3 \sin (3 x - y)\\\frac {\partial^ {2} U}{\partial x ^ {2}} = - 9 \cos (3 x - y)\\\frac {\partial U}{\partial y} = \sin (3 x - y)\\\frac {\partial^ {2} U}{\partial y ^ {2}} = - \cos (3 x - y)\end{array}\right. U ( x , y ) = cos ( 3 x − y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = − 3 sin ( 3 x − y ) ∂ x 2 ∂ 2 U = − 9 cos ( 3 x − y ) ∂ y ∂ U = sin ( 3 x − y ) ∂ y 2 ∂ 2 U = − cos ( 3 x − y )
Then,
{ ∂ 2 U ∂ y 2 = − cos ( 3 x − y ) 9 ⋅ ∂ 2 U ∂ x 2 = 9 ⋅ [ − 9 cos ( 3 x − y ) ] = − 81 cos ( 3 x − y ) → ∂ 2 U ∂ y 2 ≠ 9 ⋅ ∂ 2 U ∂ x 2 \left\{ \begin{array}{c} \frac {\partial^ {2} U}{\partial y ^ {2}} = - \cos (3 x - y) \\ 9 \cdot \frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \cdot [ - 9 \cos (3 x - y) ] = - 8 1 \cos (3 x - y) \end{array} \right. \to \frac {\partial^ {2} U}{\partial y ^ {2}} \neq 9 \cdot \frac {\partial^ {2} U}{\partial x ^ {2}} { ∂ y 2 ∂ 2 U = − cos ( 3 x − y ) 9 ⋅ ∂ x 2 ∂ 2 U = 9 ⋅ [ − 9 cos ( 3 x − y )] = − 81 cos ( 3 x − y ) → ∂ y 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U
Conclusion,
U ( x , y ) = cos ( 3 x − y ) − is not solution of the given equation U (x, y) = \cos (3 x - y) - \text{is not solution of the given equation} U ( x , y ) = cos ( 3 x − y ) − is not solution of the given equation
2. U ( x , y ) = x 2 + y 2 U(x,y) = x^{2} + y^{2} U ( x , y ) = x 2 + y 2
U ( x , y ) = x 2 + y 2 → { ∂ U ∂ x = 2 x ∂ 2 U ∂ x 2 = 2 ∂ U ∂ y = 2 y ∂ 2 U ∂ y 2 = 2 U(x, y) = x^{2} + y^{2} \rightarrow \left\{ \begin{array}{l} \frac{\partial U}{\partial x} = 2x \\ \frac{\partial^{2}U}{\partial x^{2}} = 2 \\ \frac{\partial U}{\partial y} = 2y \\ \frac{\partial^{2}U}{\partial y^{2}} = 2 \end{array} \right. U ( x , y ) = x 2 + y 2 → ⎩ ⎨ ⎧ ∂ x ∂ U = 2 x ∂ x 2 ∂ 2 U = 2 ∂ y ∂ U = 2 y ∂ y 2 ∂ 2 U = 2
Then,
{ ∂ 2 U ∂ y 2 = 2 9 ⋅ ∂ 2 U ∂ x 2 = 9 ⋅ 2 = 18 → ∂ 2 U ∂ y 2 = 2 ≠ 18 = 9 ⋅ ∂ 2 U ∂ x 2 \left\{ \begin{array}{c} \frac{\partial^{2}U}{\partial y^{2}} = 2 \\ 9 \cdot \frac{\partial^{2}U}{\partial x^{2}} = 9 \cdot 2 = 18 \end{array} \right. \rightarrow \frac{\partial^{2}U}{\partial y^{2}} = 2 \neq 18 = 9 \cdot \frac{\partial^{2}U}{\partial x^{2}} { ∂ y 2 ∂ 2 U = 2 9 ⋅ ∂ x 2 ∂ 2 U = 9 ⋅ 2 = 18 → ∂ y 2 ∂ 2 U = 2 = 18 = 9 ⋅ ∂ x 2 ∂ 2 U
Conclusion,
U ( x , y ) = x 2 + y 2 − is not solution of the given equation U(x,y) = x^{2} + y^{2} - \text{is not solution of the given equation} U ( x , y ) = x 2 + y 2 − is not solution of the given equation
3. U ( x , y ) = sin ( 3 x − y ) U(x,y) = \sin(3x - y) U ( x , y ) = sin ( 3 x − y )
U ( x , y ) = sin ( 3 x − y ) → { ∂ U ∂ x = 3 cos ( 3 x − y ) ∂ 2 U ∂ x 2 = − 9 sin ( 3 x − y ) ∂ U ∂ y = − cos ( 3 x − y ) ∂ 2 U ∂ y 2 = − sin ( 3 x − y ) U(x, y) = \sin(3x - y) \rightarrow \left\{ \begin{array}{l} \frac{\partial U}{\partial x} = 3\cos(3x - y) \\ \frac{\partial^{2}U}{\partial x^{2}} = -9\sin(3x - y) \\ \frac{\partial U}{\partial y} = -\cos(3x - y) \\ \frac{\partial^{2}U}{\partial y^{2}} = -\sin(3x - y) \end{array} \right. U ( x , y ) = sin ( 3 x − y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = 3 cos ( 3 x − y ) ∂ x 2 ∂ 2 U = − 9 sin ( 3 x − y ) ∂ y ∂ U = − cos ( 3 x − y ) ∂ y 2 ∂ 2 U = − sin ( 3 x − y )
Then,
{ ∂ 2 U ∂ y 2 = − sin ( 3 x − y ) 9 ⋅ ∂ 2 U ∂ x 2 = 9 ⋅ [ − 9 sin ( 3 x − y ) ] = − 81 sin ( 3 x − y ) → ∂ 2 U ∂ y 2 ≠ 9 ⋅ ∂ 2 U ∂ x 2 \left\{ \begin{array}{c} \frac{\partial^{2}U}{\partial y^{2}} = -\sin(3x - y) \\ 9 \cdot \frac{\partial^{2}U}{\partial x^{2}} = 9 \cdot [-9\sin(3x - y)] = -81\sin(3x - y) \end{array} \right. \rightarrow \frac{\partial^{2}U}{\partial y^{2}} \neq 9 \cdot \frac{\partial^{2}U}{\partial x^{2}} { ∂ y 2 ∂ 2 U = − sin ( 3 x − y ) 9 ⋅ ∂ x 2 ∂ 2 U = 9 ⋅ [ − 9 sin ( 3 x − y )] = − 81 sin ( 3 x − y ) → ∂ y 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U
Conclusion,
U(x,y) = \sin (3x - y) - \text{is not solution of the given equation}
4.A. U ( x , y ) = e − 3 π x ⋅ sin ( π y ) U(x,y) = e^{-3\pi x}\cdot \sin (\pi y) U ( x , y ) = e − 3 π x ⋅ sin ( π y )
U ( x , y ) = e − 3 π x ⋅ sin ( π y ) → { ∂ U ∂ x = − 3 π ⋅ e − 3 π x ⋅ sin ( π y ) ∂ 2 U ∂ x 2 = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ U ∂ y = π ⋅ e − 3 π x ⋅ cos ( π y ) ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) U (x, y) = e ^ {- 3 \pi x} \cdot \sin (\pi y) \rightarrow \left\{\begin{array}{l}\frac {\partial U}{\partial x} = - 3 \pi \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y)\\\frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y)\\\frac {\partial U}{\partial y} = \pi \cdot e ^ {- 3 \pi x} \cdot \cos (\pi y)\\\frac {\partial^ {2} U}{\partial y ^ {2}} = - \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y)\end{array}\right. U ( x , y ) = e − 3 π x ⋅ sin ( π y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = − 3 π ⋅ e − 3 π x ⋅ sin ( π y ) ∂ x 2 ∂ 2 U = 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ y ∂ U = π ⋅ e − 3 π x ⋅ cos ( π y ) ∂ y 2 ∂ 2 U = − π 2 ⋅ e − 3 π x ⋅ sin ( π y )
Then,
{ ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) 9 ⋅ ∂ 2 U ∂ x 2 = 9 ⋅ [ 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ] = 81 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) → \left\{ \begin{array}{c} \frac {\partial^ {2} U}{\partial y ^ {2}} = - \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) \\ 9 \cdot \frac {\partial^ {2} U}{\partial x ^ {2}} = 9 \cdot [ 9 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) ] = 8 1 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) \end{array} \right. \to { ∂ y 2 ∂ 2 U = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) 9 ⋅ ∂ x 2 ∂ 2 U = 9 ⋅ [ 9 π 2 ⋅ e − 3 π x ⋅ sin ( π y )] = 81 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) → ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ≠ 81 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) = 9 ⋅ ∂ 2 U ∂ x 2 \frac {\partial^ {2} U}{\partial y ^ {2}} = - \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) \neq 8 1 \pi^ {2} \cdot e ^ {- 3 \pi x} \cdot \sin (\pi y) = 9 \cdot \frac {\partial^ {2} U}{\partial x ^ {2}} ∂ y 2 ∂ 2 U = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) = 81 π 2 ⋅ e − 3 π x ⋅ sin ( π y ) = 9 ⋅ ∂ x 2 ∂ 2 U
Conclusion,
U(x,y) = e^{-3\pi x} \cdot \sin(\pi y) - \text{is not solution of the given equation}
4.B. U ( x , y ) = e − 3 π x ⋅ sin ( π y ) U(x,y) = e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) U ( x , y ) = e − π x 3 ⋅ sin ( π y )
U ( x , y ) = e − 3 π x ⋅ sin ( π y ) → { ∂ U ∂ x = 3 π x 2 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ 2 U ∂ x 2 = − 6 π x 3 ⋅ e − 3 π x ⋅ sin ( π y ) + 9 π 2 ⋅ x 4 ⋅ e − 3 π x ⋅ sin ( π y ) ∂ U ∂ y = π ⋅ e − 3 π x ⋅ cos ( π y ) ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) U(x,y) = e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \rightarrow \left\{ \begin{array}{c}
\frac{\partial U}{\partial x} = \frac{3}{\pi x^2} \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \\
\frac{\partial^2 U}{\partial x^2} = -\frac{6}{\pi x^3} \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) + \frac{9}{\pi^2 \cdot x^4} \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \\
\frac{\partial U}{\partial y} = \pi \cdot e^{-\frac{3}{\pi x}} \cdot \cos(\pi y) \\
\frac{\partial^2 U}{\partial y^2} = -\pi^2 \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y)
\end{array} \right. U ( x , y ) = e − π x 3 ⋅ sin ( π y ) → ⎩ ⎨ ⎧ ∂ x ∂ U = π x 2 3 ⋅ e − π x 3 ⋅ sin ( π y ) ∂ x 2 ∂ 2 U = − π x 3 6 ⋅ e − π x 3 ⋅ sin ( π y ) + π 2 ⋅ x 4 9 ⋅ e − π x 3 ⋅ sin ( π y ) ∂ y ∂ U = π ⋅ e − π x 3 ⋅ cos ( π y ) ∂ y 2 ∂ 2 U = − π 2 ⋅ e − π x 3 ⋅ sin ( π y )
Then,
{ ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) 9 ⋅ ∂ 2 U ∂ x 2 = 9 ⋅ [ ( 9 π 2 x 4 − 6 π x 3 ) ⋅ e − 3 π x ⋅ sin ( π y ) ] = ( 81 π 2 x 4 − 54 π x 3 ) ⋅ e − 3 π x ⋅ sin ( π y ) \left\{ \begin{array}{c}
\frac{\partial^2 U}{\partial y^2} = -\pi^2 \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \\
9 \cdot \frac{\partial^2 U}{\partial x^2} = 9 \cdot \left[ \left(\frac{9}{\pi^2 x^4} - \frac{6}{\pi x^3}\right) \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \right] = \left(\frac{81}{\pi^2 x^4} - \frac{54}{\pi x^3}\right) \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y)
\end{array} \right. { ∂ y 2 ∂ 2 U = − π 2 ⋅ e − π x 3 ⋅ sin ( π y ) 9 ⋅ ∂ x 2 ∂ 2 U = 9 ⋅ [ ( π 2 x 4 9 − π x 3 6 ) ⋅ e − π x 3 ⋅ sin ( π y ) ] = ( π 2 x 4 81 − π x 3 54 ) ⋅ e − π x 3 ⋅ sin ( π y ) ∂ 2 U ∂ y 2 = − π 2 ⋅ e − 3 π x ⋅ sin ( π y ) ≠ ( 81 π 2 x 4 − 54 π x 3 ) ⋅ e − 3 π x ⋅ sin ( π y ) = 9 ⋅ ∂ 2 U ∂ x 2 \frac{\partial^2 U}{\partial y^2} = -\pi^2 \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) \neq \left(\frac{81}{\pi^2 x^4} - \frac{54}{\pi x^3}\right) \cdot e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) = 9 \cdot \frac{\partial^2 U}{\partial x^2} ∂ y 2 ∂ 2 U = − π 2 ⋅ e − π x 3 ⋅ sin ( π y ) = ( π 2 x 4 81 − π x 3 54 ) ⋅ e − π x 3 ⋅ sin ( π y ) = 9 ⋅ ∂ x 2 ∂ 2 U
Conclusion,
U ( x , y ) = e − 3 π x ⋅ sin ( π y ) − is not solution of the given equation \boxed{U(x,y) = e^{-\frac{3}{\pi x}} \cdot \sin(\pi y) - \text{is not solution of the given equation}} U ( x , y ) = e − π x 3 ⋅ sin ( π y ) − is not solution of the given equation ANSWER
i f ∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ x 2 → ∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ y 2 if \quad \frac{\partial^2 U}{\partial x^2} = 9 \cdot \frac{\partial^2 U}{\partial x^2} \rightarrow \frac{\partial^2 U}{\partial x^2} = 9 \cdot \frac{\partial^2 U}{\partial y^2} i f ∂ x 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U → ∂ x 2 ∂ 2 U = 9 ⋅ ∂ y 2 ∂ 2 U
Then, solutions are
1. U ( x , y ) = cos ( 3 x − y ) U(x,y) = \cos(3x - y) U ( x , y ) = cos ( 3 x − y ) U ( x , y ) = sin ( 3 x − y ) U(x,y) = \sin(3x - y) U ( x , y ) = sin ( 3 x − y )
i f ∂ 2 U ∂ x 2 = 9 ⋅ ∂ 2 U ∂ x 2 → ∂ 2 U ∂ y 2 = 9 ⋅ ∂ 2 U ∂ x 2 if \quad \frac{\partial^2 U}{\partial x^2} = 9 \cdot \frac{\partial^2 U}{\partial x^2} \rightarrow \frac{\partial^2 U}{\partial y^2} = 9 \cdot \frac{\partial^2 U}{\partial x^2} i f ∂ x 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U → ∂ y 2 ∂ 2 U = 9 ⋅ ∂ x 2 ∂ 2 U
Then, solutions are
1. U ( x , y ) = cos ( 3 x − y ) U(x,y) = \cos (3x - y) U ( x , y ) = cos ( 3 x − y ) U ( x , y ) = sin ( 3 x − y ) U(x,y) = \sin (3x - y) U ( x , y ) = sin ( 3 x − y )
None of the presented variants is a solution of this equation
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#339914 on Dec 2023